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Andreyy89
3 years ago
10

The cloth shroud from around a mummy is found to have a 14C activity of 8.4 disintegrations per minute per gram of carbon as com

pared with living organisms that undergo 15.2 disintegrations per minute per gram of carbon.From the half-life for 14C decay, 5715 yr, calculate the age of the shroud
Chemistry
1 answer:
kondaur [170]3 years ago
8 0

Answer:

48889 years

Explanation:

Activity of the living sample Ao= 15.2 disintegration per minute

Activity of the cloth A= 8.4 disintegration per minute

Half life of 14C= 5715 years

Time taken for the decay t= the unknown

From;

0.693/5715 = 2.303/t log (15.2/8.4)

1.213 × 10^-4 = 2.303/t log (1.8095)

1.213 × 10^-4 = 0.593/t

t= 0.593/1.213 × 10^-4

t= 4888.7 years

t= 48889 years

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Answer:

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b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

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