Radioactive decay => C = Co { e ^ (- kt) |
Data:
Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min
Time conversion: 4 hr 39 min = 4.65 hr
1) Replace the data in the equation to find k
C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}
=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719
2) Use C / Co = 1/2 to find the hallf-life
C / Co = e ^ (-kt) => -kt = ln (C / Co)
=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k
t = ln(2) / 0.44719 = 1.55 hr.
Answer: 1.55 hr
No, because hydrogen isn’t brought out of the equation
Answer:
Element with 6s subshell
Explanation:
Reactivity of an element depends on the electronic configuration and position of element in the periodic table as reactivity increases as we go down the periodic table.
This is so because number of shell increases as move down the periodic table and the last electron is further away from the nucleus.
Element with 6s subshell is the largest among 3s and 4s subshell and has more number of shells so it will react more than 3s and 4s subshell.
Hence, the correct answer is "Element with 6s subshell".
