Answer:
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Explanation:
Given data:
Atomic mass of silicon= ?
Percent abundance of Si-28 = 92.21%
Atomic mass of Si-28 = 27.98 amu
Percent abundance of Si-29 = 4.70%
Atomic mass of Si-29 = 28.98 amu
Percent abundance of Si-30 = 3.09%
Atomic mass of Si-30 = 29.97 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100
Average atomic mass = 2580.04 +136.21+92.61 / 100
Average atomic mass = 2808.86 / 100
Average atomic mass = 28.08amu.
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Hello!
On the periodic table, as we go down the periodic table, the ionization energy decreases, but as we go across the periodic table (left to right), the ionization increases.
On the periodic table, lithium (Li) is located in column one, beryllium (Be) is located in column two, and (B) boron is located in column 13. As stated above, when we go across the periodic table (left to right), the ionization increases.
Therefore, the element with the highest ionization energy is Boron, or symbol B on the period table.
Hahahaaaa none of the above but IF <span>(c) is
1/2 mole of NaCl and 1/3 mole of MgCl2 instead,
then C is the right ans :)</span>
The paper is not clear so please ask your problem again with more clear print
