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.
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Answer: definite proportions.
Explanation:
1) The definite proportions law states that compounds will always have the same kind of atoms (elements) in the same mass proportion (ratios).
2) For example, a molecule of water will alwys have the same mass ratio of hydrogen atoms to oxygen atoms. That is what permits to obtain the chemical formula of the water molecule as H₂O.
The mass of the two hydrogen atoms will be in a fixed ratio respect to the mass of the oxygen atoms.
Then, if you have one reactant in less proportion than the other, respect to the ratio stated by the chemical formula of water, the former will react completely (it is the limiting reactant) with the corresponding (proportional) mass of the later. Then there will be an excess of the later reactant which will not react (will remain unchanged).
The reactants can only react in the proportion defined by the chemical formulas of the final products.
Oxidation state of I is (-1) and for CO it is zero. Let's assume that the oxidation state of Fe in Fe(CO)₄I₂<span> (s) is x. For whole compound, the charge is zero.
Sum of oxidation numbers in all elements = Charge of the compound.
Here we have 1Fe , 4CO and 2I
hence we can find the oxidation state as;
x + 4*0 + 2*(-1) = 0
x + 0 - 2 = 0
x = +2
Hence the oxidation state of Fe in product </span>Fe(CO)₄I₂ (s) is +2.
Same as we can find the oxidation state (y) of Fe in Fe(CO)₅(s).
y + 5*0 = 0
y = 0
Since oxidation state of Fe increased from 0 to +2, the oxidized element is Fe in the given reaction.
20 mol of NH, can be produce from 30 mol o H2
