Explanation:
The given chemical reaction is:
The relation between Eo cell and Keq is shown below:
The value of Eo cell is:
Br- undergoes oxidation and I2 undergoes reduction.
Reduction takes place at cathode.
Oxidation takes place at anode.
Hence,
F=96485 C/mol
n=2 mol
R=8.314 J.K-1.mol-1
T=298K
Substitute all these values in the above formula:
Answer:
Keq=6.13x10^33
The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.
