A wooden artifact from a Chinese temple has a^14C activity of 32.0 counts per mi

Question

A wooden artifact from a Chinese temple has a^14C activity of 32.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. The half-life for^14C decay is 5715 year. Determine the age of the artifact. In some applications, nickel-cadmium batteries (EMF = 1.30 V) have been replaced by nickel zinc batteries. The overall cell reaction for this new battery is: 2H_2O(l) + 2NiO(OH)(s) + Zn(s) implies 2Ni(OH)_2(s) + Zn(OH)_2(s) What is the cathode half reaction? What is the anode half reaction? What voltage would you expect the nickel-zinc battery will produce?

Explanation / Answer

A = 32

A0 = 58.2 at t = 0

HL = 5715

determine age

apply half life eqution

A = A0*(1/2)^(t/HL)

32 = 58.2 *(1/2)^(t/5715)

ln(32/58.2) / ln(1/2) = t/5715)

t = 0.8629*5715

t = 4931.4735 years old

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