In C.AJ -Kt + ln [A.o]. t1/₂ = 0.693 K 'In = KE CAJO. CAJE EAJo 100./. CAJO [A] += left it = ln [AG] K [100] [s]t .In. K= 0.693 t11₂ =kt CAJO K= In [AE]. t Formulas Problem: Na-24 has a half life of 15 hours. How much of a 20.09. Sample would remain after decaying for 60 hours? Plz explain step by step using the formulas..

Introductory Chemistry: A Foundation
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Chapter15: Solutions
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Problem 106AP: A certain grade of steel is made by dissolving 5.0 g of carbon and 1.5 g of nickel per 100. g of...
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. In C.AJ = - kt.tln [A.o].
t¹/₂ = 0.693
K
In
= KE
•1• [A]o⋅=⋅· 100 ·/.
[A] + = left
.
CAJO.
CAJE
In.
[100]
[s] t
=.kt
K= 0.693
t¼½₂
CAJO
t = |n² Ch
t = ln [AG]
·K
CAJO
K= In [At].
t
Formulas
Problem:
Na-24 has a half life of 15 hours. How much of a 20.09.
·Sample would remain after decaying for 60 hours?
Plz explain step by step. using the formulas..
Transcribed Image Text:. In C.AJ = - kt.tln [A.o]. t¹/₂ = 0.693 K In = KE •1• [A]o⋅=⋅· 100 ·/. [A] + = left . CAJO. CAJE In. [100] [s] t =.kt K= 0.693 t¼½₂ CAJO t = |n² Ch t = ln [AG] ·K CAJO K= In [At]. t Formulas Problem: Na-24 has a half life of 15 hours. How much of a 20.09. ·Sample would remain after decaying for 60 hours? Plz explain step by step. using the formulas..
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