Asked by Anonymous
                the half-life of radon-222 is 3.823 days. what was the original mass if 0.050g remains after 7.646 days?
            
            
        Answers
                    Answered by
            DrBob222
            
    k= 0.693/t<sub>1/2</sub>
Then substitute k into the equation below.
ln(No/N) = kt
Solve for No.
N = 0.05g
k = from above.
t = 7.646 days.
    
Then substitute k into the equation below.
ln(No/N) = kt
Solve for No.
N = 0.05g
k = from above.
t = 7.646 days.
                    Answered by
            Anonymous
            
    The half-life of radon-222 is 3.823 days. What was the orginal mass if 0.050g reamains after 7.646 days? Show your work 
    
                    Answered by
            ABDULLA
            
    236
    
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