ForFinal_AIK

ForFinal_AIK - Material for the final exam 1/3: new...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Material for the final exam 1/3: new material (transition metals, nuclear chemistry) 2/3: old material (see exams 1-4 for the coverage, review keys !) Total: ~12 problems. Some problems may combine concepts from different chapters. No special topics, no graph paper problems Today: Transition metals and nuclear chemistry, some thermodynamics
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The mass rations of 40 Ar and 40 K can be used to date geological materials. 40 K decays by two processes: 40 19 K + 0 -1 e -> 40 18 Ar (10.7%) 40 19 K -> 40 20 Ca + 0 1 e (89.3%) Overall half-life of 40 K is 1.27x10 9 years. (1) What assumptions must be made in using this technique? (2) A sedimentary rock has a 40 Ar/ 40 K ratio of 0.95. Calculate the age of the rock. (3) How will the measured age of a rock compare with the actual age if some 40 Ar has escaped from the sample? 1. Assumptions: (i) no Ar escapes the rock; (ii) 40 K is the sole source of 40 Ar. 2. Use: t t t t kt N N 2 / 1 2 / 1 0 693 . 0 ) 2 ln( ) ln( 693 . 0 ) ln( 693 . 0 ) ln( 2 / 1 0 2 / 1 0 t N N t N N t 3. Determine N 0 /N . N current amount of 40 K, N 0 initial amount of 40 K. 10.7% N 0 -> decomposes through channel 1, producing 40 Ar ( N Ar ) 100% N 0 -> decomposes through both cannels ( N d ) N d =100/10.7*N Ar =9.35*N Ar : total amount of 40 K decomposed
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/22/2008 for the course CHEM 115b taught by Professor Reisler during the Spring '08 term at USC.

Page1 / 7

ForFinal_AIK - Material for the final exam 1/3: new...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online