Answer:
A long uniformly charged wire has charge density λ=0.16μλ=0.16μC/m.
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.
Answer:
Option 4 with o-h in the most polar bond, since the two atoms in the bond have the greatest difference in electronegativity. This is assuming there are no other factors in other atoms bound to either of the elements in the bond.
Explanation:
The building blocks of protein are amino acids.
Amino acids are a class of organic compounds that contain at least one amino group, -NH2, and carboxyl group, -COOH.
Alpha amino acids , RCH(NH2)COOH, are the building blocks from which proteins are constructed.
The NH group of one amino acid and the COOH group of the other amino acid are joined together and a peptide bond -CONH- is formed between the two amino acids and the product is called a dipeptide.
D. amu
amu stands for atomic mass unit
