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How many Coinium atoms would remain after three half-lives, if the original sample had 250 Coinium

atoms? Show your work.

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It seems there might be a mistake in your question. "Coinium" is not a real element or is not recognized in the context of atomic decay. If you meant a different element, please provide the correct name. Assuming you meant a generic element with a half-life, I can provide you with a general formula for radioactive decay. The formula for calculating the remaining amount of a substance after a certain number of half-lives is given by: � = � 0 × ( 1 2 ) � N=N 0 ​ ×( 2 1 ​ ) n Where: � N is the final amount of the substance, � 0 N 0 ​ is the initial amount of the substance, � n is the number of half-lives. In your case, � 0 = 250 N 0 ​ =250 atoms and � = 3 n=3 half-lives. Plug these values into the formula: � = 250 × ( 1 2 ) 3 N=250×( 2 1 ​ ) 3 Calculate each step: � = 250 × ( 1 2 ) × ( 1 2 ) × ( 1 2 ) N=250×( 2 1 ​ )×( 2 1 ​ )×( 2 1 ​ ) � = 250 × 1 8 N=250× 8 1 ​ � = 31.25 N=31.25 So, after three half-lives, approximately 31.25 atoms of the substance would remain. Note that you cannot have a fraction of an atom in reality, so you would round to the nearest whole number. Therefore, in this case, after three half-lives, approximately 31 atoms of the substance would remain.

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