A. The initial mass of the sample is 6.35
B. The time taken for 0.3 mg of the sample to remain is 66 days
A. Determination of the initial mass of the sample
Half-life (t½) = 15 days
Time (t) = 10 days
n = t / t½
n = 10 / 15
Number of half-lives (n) = ⅔
Amount remaining (N) = 4 mg
[tex]N_0 = N \times {2}^{n} \\ N_0 = 4 \times {2}^{2 \div 3} \\ N_0 = 6.35 \: mg[/tex]
Therefore, the initial mass of the sample is 6.35 mg
B. Determination of the time
Amount remaining (N) = 0.3 mg
Initial amount (N₀) = 6.35 mg
[tex]{2}^{n} = \frac{N_0}{N} \\ \\ {2}^{n} = \frac{6.35}{0.3} \\ \\ {2}^{n} = 21.167 \\ take \: the \: log \: of \: both \: side \\ log \: {2}^{n} = log \: 21.167 \\ nlog2 = log \: 21.167 \\ divide \: both \: side \: by \: log2 \\ \\ n = \frac{log \: 21.167}{log2} \\ \\ n = 4.4[/tex]
Number of half-lives (n) = 4.4
Half-life (t½) = 15 days
t = n × t½
t = 4.4 × 15
Therefore, it will take 66 days for 0.3 mg of the sample to remain.
Learn more: https://brainly.com/question/25749387