The Average atomic weight of X is 28.7amu
Isotopes are atoms with the same number of protons but differing numbers of neutrons.
Different isotopes have various atomic masses.
The proportion of atoms with a particular atomic mass that can be found in a naturally occurring sample of an element is known as the relative abundance of an isotope.
An element's average atomic mass is computed as a weighted average by multiplying the relative abundances of its isotopes by their respective atomic masses, then adding the resulting products.
Using mass spectrometry, it is possible to determine the relative abundance of each isotope.
The atomic weight of the element will be a weighted average of the isotopes based on the relative abundance:
(27.730 x 0.6058) + (28.841 x 0.1835) + (31.321 x 0.2107) = 16.7988 + 5.2923+ 6.599 = 28.690 = 28.7 amu.
Average atomic weight of X is 28.7amu
Learn more about Average Atomic Weight here
brainly.com/question/6200158
#SPJ4
Answer:
mostly gasoline in cars nowadays
The nuclei of atoms also contain neutrons, which help hold the nucleus together. ... The total weight of an atom is called the atomic weight. It is approximately equal to the number of protons and neutrons, with a little extra added by the electrons.
The mass of a given atom, measured on a scale in which the hydrogen atom has the weight of one. Because most of the mass in an atom is in the nucleus, and each proton and neutron has an atomic weight near one, the atomic weight is very nearly equal to the number of protons and neutrons in the nucleus.
Answer : The percentage reduction in intensity is 79.80 %
Explanation :
Using Beer-Lambert's law :



where,
A = absorbance of solution
C = concentration of solution = 
l = path length = 2.5 mm = 0.25 cm
= incident light
= transmitted light
= molar absorptivity coefficient = 
Now put all the given values in the above formula, we get:



If we consider
= 100
then, 
Here 'I' intensity of transmitted light = 20.198
Thus, the intensity of absorbed light
= 100 - 20.198 = 79.80
Now we have to calculate the percentage reduction in intensity.


Therefore, the percentage reduction in intensity is 79.80 %