Answer:
Polyunsaturated
Explanation:
A fatty acid is a compound that is derivative from a carboxylic acid. The chain of carbons can have since 4 carbons to more than 20 carbons. The bonds between carbons can be simple, which are called saturated bonds, or double, or triple, which is called unsaturated bonds.
When a fatty acid only has saturated bonds between carbons, it's called a saturated fatty acid. If there is only one unsaturated bond, it's called an unsaturated fatty acid, and if there are two or more unsaturated bonds, it's called polyunsaturated fatty acid.
Answer:
![r=k[UO_2^+]^2[H^+]](https://tex.z-dn.net/?f=r%3Dk%5BUO_2%5E%2B%5D%5E2%5BH%5E%2B%5D)
Explanation:
The given reaction is :-
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
m = 2 = is order with respect to 
n = order with respect to H+
overall order = m+n = 3
n = 3 - m = 3 - 2 = 1
Rate law is:-
Explanation:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
The atomic mass of isotope (II) of vanadium is 49.944 amu.
<span>A total mechanical energy of an object is calculated as the sum of kinetic energy and potential energy</span>
