Suppose the amount of a certain radioactive substance in a sample decays from to over a period of days. Calculate the half life of the substance. Round your answer to significant digit.

The given question is incomplete, the complete question is:

Suppose the amount of a certain radioactive substance in a sample decays from 1.30 mg to 100. ug over a period of 29.5 days. Calculate the half life of the substance Round your answer to 2 significant digits.

The correct answer is 7.974 days.

Explanation:

Based on the given question, the concentration of a radioactive substance present in a sample get decays to 100 micro grams from 1.30 milligrams in 29.5 days. There is a need to find the half-life of the substance.  

Radioactive decay is an illustration of first order reaction.  

K = (2.303 / t) log [a/(a-x)]

Here a is 1.30 mg and (a-x) is 100 micrograms = 100 * 10^-3 mg or 0.1 mg, and t is 29.5 days. Now putting the values we get,  

K = (2.303 /29.5)log (1.30/0.1)

= 2.303/29.5 log13

= 2.303/29.5 * 1.1139

K = 0.0869

The half-life or t1/2 is calculated by using the formula, 0.693 / K

= 0.693 / 0.0869

= 7.974 days.  

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.