Answer:
Relative abundance of each isotopes of an element.
Explanation:
Consider the following example:
Uranium is used in nuclear reactors and is a rare element on earth. Uranium has three common isotopes. If the abundance of 234U is 0.01% the abundance of 235U is 0.71% and the abundance of 238U is 99.28% what is the average atomic mass of uranium
Abundance of U²³⁴ = 0.01%
Abundance of U²³⁵ = 0.17%
Abundance of U²³⁸ = 99.28%
Average atomic mass = ?
Solution:
Average atomic mass of uranium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) +(abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass of uranium= (234×0.01)+(235×0.71)+(238×99.28)/100
Average atomic mass of uranium= 2.34 + 166.85 + 23628.64 / 100
Average atomic mass of uranium= 23797.83 / 100
Average atomic mass of uranium= 237.98 amu.
Answer:
b) F
Explanation:
Only non-metals can gain electrons during ionic bonding.
b) F
Li, Mg and K are metals.
<span>Fe+ PbSO</span>₂<span>-->FeSO</span>₂<span>+2Pb</span>
The answer is option C 7.60 M
Answer:
Energy can change form, so False
