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.
The correct answer is (D)
All the above are reasons the carbon , oxygen, and nitrogen cycles are vital to life on earth.
The explanation:
because :
1) Carbon, oxygen, and nitrogen are vital components of life on Earth.
2) The carbon, oxygen, and nitrogen cycles allow vital elements to return to usable form by organisms.
3)The carbon, oxygen, and nitrogen cycles are an important interface between biotic and abiotic factors
4) They are all biogeochemical cycles.
5) They all involve an interaction between living and nonliving elements.
6)They are all part of the Earth system.
No i think they can not be
When pcl5 solidifies it forms pcl4 cations and pcl6–anions. according to valence bond theory, hybrid orbitals that are used by phosphorus in the pcl4 cations are one orbital of s and three orbital of p as it is sp³hydridised.
<h3>What is sp³ hybridization?</h3>
Hybridization is a process or system which specifies the shape and geometry of any element or molecule with bond angles too.
The pcl4 cation is sp³ hybridized because of the phosphorus behave as a central atom here and the 4 chloride molecules are attached with the p- orbitals to the phosphorus molecule.
Therefore, pcl4 cation is sp³ hybridized.
Learn more about sp³ hybridization, here :
brainly.com/question/13062274
#SPJ4
Answer:
1.71x10²⁷
Explanation:
If we sum 1/2 of (3) + 1/2 of (1):
1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷ = 4.58x10²³
1/2 (1) 1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8
C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>
K' = 4.58x10²³ * 11.8 = 5.42x10²⁴
+1/2 (2):
<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2
C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)
K'' = 5.42x10²⁴* 316.2 =
<h3>1.71x10²⁷</h3>