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):
The power needed to process 50 ton/hor is
2) The density of the packed bed can be expressed as
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
The void fraction of the bed is 0.37.
Balanced equation for the above reaction is as follows;
Mg(OH)₂ + 2HCl ---> MgCl₂ + 2H₂O
stoichiometry of Mg(OH)₂ to MgCl₂ is 1:1
mass of Mg(OH)₂ reacted - 1.82 g
number of moles of Mg(OH)₂ - 1.82 g/ 58.3 g/mol = 0.0312 mol
number of Mg(OH)₂ moles reacted - number of MgCl₂ moles formed
number of MgCl₂ moles formed - 0.0312 mol
mass of MgCl₂ formed - 0.0312 mol x 95.2 g/mol = 2.97 g
mass of MgCl₂ formed - 2.97 g
Answer:
It will have 5 valence electrons as its in group 5.
The atom will gain as its closer to the full configuration 8.
The charge will be 3- as it will gain 3 electrons.
Equation of decomposition of ammonia:
N2+3H2->2NH3
Euilibrium constant:
Kc=(NH3)^2/((N2)((H2)^3))
As concentration of N2=0.000105, H2=0.0000542
so equation will become:
3.7=(NH3)^2/(0.000105)*(0.0000542)^3
NH3=√(3.7*0.000105*(0.0000542)^3)
NH3=7.8×10⁻⁹
So concentration of ammonia will be 7.8×10⁻⁹.
