Bromine-88 is radioactive and has a half life of 16.3 seconds. what percentage of a sample would be left after 35.0 seconds? round your answer to 2 significant digits .

Percentage of bromine-88 left after 35.0 s = 23%

Explanation:

Radioactive decay follows first order kinetics.

Given:

Half life of Br-88 = 16.3 s

time = 35.0 s

where,

= Initial concentration of radioactive substance

a = Amount left after time 't'

K = rate constant

Half-life = 0.693/K

now, substitute the value of rate constant and t (35.0 s) in the formula,

Let a0 (initial concentration of bromine-88) be 100.

[a] = 22.6% = 23%

Percentage of bromine-88 left after 35.0 s = 23%

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

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