Answer :
staals half
1 gram left
0.3 grams left
half lives are normally written in years
it goes likt this
[tex]A=P( \frac{1}{2})^ \frac{t}{h} [/tex]
A=final amount
P=initila amount
t=time elapsed
h=time half life is (same units as time)
let's use days as the units of time
we are given
initial amount=1gram
final amount=0.3gram
time=4 days
so
[tex]0.3=1( \frac{1}{2})^ \frac{4}{h} [/tex]
solve for h
[tex]0.3=( \frac{1}{2})^ \frac{4}{h} [/tex]
take the ln of both sides
[tex]ln(0.3)=ln( \frac{1}{2})^ \frac{4}{h} [/tex]
[tex]ln(0.3)=\frac{4}{h}ln( \frac{1}{2}) [/tex]
times both sides by h
[tex]hln(0.3)=4ln( \frac{1}{2}) [/tex]
divide both sides by ln(0.3)
[tex]h= \frac{4ln( \frac{1}{2}) }{ln(0.3)} [/tex]
use calculator
h=2.30287
3 sig figs
h=2.303 days
1 gram left
0.3 grams left
half lives are normally written in years
it goes likt this
[tex]A=P( \frac{1}{2})^ \frac{t}{h} [/tex]
A=final amount
P=initila amount
t=time elapsed
h=time half life is (same units as time)
let's use days as the units of time
we are given
initial amount=1gram
final amount=0.3gram
time=4 days
so
[tex]0.3=1( \frac{1}{2})^ \frac{4}{h} [/tex]
solve for h
[tex]0.3=( \frac{1}{2})^ \frac{4}{h} [/tex]
take the ln of both sides
[tex]ln(0.3)=ln( \frac{1}{2})^ \frac{4}{h} [/tex]
[tex]ln(0.3)=\frac{4}{h}ln( \frac{1}{2}) [/tex]
times both sides by h
[tex]hln(0.3)=4ln( \frac{1}{2}) [/tex]
divide both sides by ln(0.3)
[tex]h= \frac{4ln( \frac{1}{2}) }{ln(0.3)} [/tex]
use calculator
h=2.30287
3 sig figs
h=2.303 days