The oxidation number of elements in equation below are,
4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂
O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1
Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).
Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).
Result:
<span>N is oxidized and Cl is reduced.</span>
I’m not sure if there was important information in the question before this one, but the answer based on the info I have is B.
The density of water is 1kg/L. Since the density of the block is less, it will float.
If the crucible wasn't covered with a lid the reactants may have produced a gas that was released into the surroundings, or mass may have been lost in the form of water vapour.
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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