Answer:
1.02x10^24 atoms
Explanation:
The coefficients (the numbers before each susbtance) in this equation tell us that for every 2 moles of hydrogen used, there will be 1 mole of tin produced. This is a fixed ratio, which means if we use 4 moles of H2, 2 moles of tin will be produced.
One mole contains Avogadro's number of atoms (6.02 x 10^23). So two moles would contain twice as many atoms as this.
2 x 6.02x10^23 = 1.02x10^24 atoms.
:)
Answer:
Average atomic mass of uranium= 237.98 amu.
Explanation:
Given data:
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.
Second one i think.......
Number 2 I believe is D and number 3 is most likely A