Explanation:
The given cell reaction is as follows.

Hence, reactions taking place at the cathode and anode are as follows.
At anode ; Oxidation-half reaction :
...... (1)
At cathode; Reduction-half reaction :
....... (2)
Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.
Therefore, net cell reaction is as follows.

Net reaction: 
Thus, we can conclude that the overall cell reaction is as follows.

Answer:
-3
Explanation:
The oxidation state or oxidation number of an atom is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.
The complex anion here is [Cr(CN)6]3-.
Now, as the oxidation state of CN or cyanide ligand is -1, and if we suppose the oxidation state of Cr to be 'x', then; x - 6 = -3 (overall charge on the anion),
so x= +3. Hence the oxidation state of Chromium in this complex hexacyanochromium (III) anion comes out to be -3.
.
Answer:
ionic compounds are made from metal and non mental elements
Explanation:
in this case it is know as an ionic compound because it contains a charge. For example, NaCl is simply a compound as it contains no charges (the charges cancel out as Cl is -1 and Na is +1)
but OH- is an ionic compound as it has a charge if -1 (O has a -2 charge and H has a +1 charge, so -2+1=-1 so OH has -1 charge)
These are called subscript number.
That is the number below the normal line of test are called subscript number.
This number indicate the indicate the number of atoms of the element present in the chemical formula.
In both of these C₆H₁₂O₆ and H₂O, the number written below the line of the text are called subscript numbers.
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.