Answer:
Explanation:
The<em> half-life </em>time of a radiactive isotope (radioisotope) is a constant value, meaning that the amount of the radioisotope that decays will be (1/2) raised to the number of half-lives passed.
Naming A₀ the initial amount to the radioisotope, you can build this table to find the amount left.
Number of half-lives amount of radiosotope left
0 A₀
1 (1/2) × A₀
2 (1/2)×(1/2)×A₀ = (1/2)² × A₀
3 (1/2)³ ×A ₀
4 (1/2)⁴ × A₀
n (1/2)ⁿ × A₀
Now calculate the number of half-lives the strontium-90 sample has passed after 100 years:
- n = 100 years / 28.1 years ≈ 3.5587
Hence, the amount of strontium-90 is:

In percent, that is:

Rounding to two significant figures, that is 8.5%.
<u>Conclusion</u>: <em>The percent of strontium-90 left after 100 yeaers is 8.5% </em>(choice number 4).
Answer:
Uh, the second one?
Explanation:
Try to restate the question please.
Answer:
A = B < D < C
C - S
Cl - Cl
F ← H
Si → O
Explanation:
The polarity of a bond increases with the increase in the difference in electronegativity. The dipole moment is represented with an arrow pointing the more electronegative atom.
A: carbon-sulfur
C - S
ΔEN = |EN(C) - EN(S)| = |2.5 - 2.5| = 0
B: chlorine - chlorine
Cl - Cl
ΔEN = |EN(Cl) - EN(Cl)| = |3.0 - 3.0| = 0
C: fluorine – hydrogen
F ← H
ΔEN = |EN(F) - EN(H)| = |4.0 - 2.1| = 1.9
D: silicon - oxygen
Si → O
ΔEN = |EN(Si) - EN(O)| = |1.8 - 3.5| = 1.7
The order of increasing polarity is A = B < D < C.