<h3>
Answer:</h3>
6.25 atoms
<h3>
Explanation:</h3>
<u>We are given</u>;
- The half life of Po-218 is 3 minutes
- Initial sample is 200 atom
- Time of decay is 15 minutes
We are required to calculate the remaining mass after decay;
Half life refers to the time taken for original amount of a radioactive sample to decay to a half.
To calculate the remaining mass we use the formula;
N = N₀ × 0.5^n where n is the number of half lives, N is the new amount and N₀ is the original amount.
n = 15 min ÷ 3 min
= 5
Therefore;
New amount = 200 atom × 0.5^5
= 6.25 atoms
Therefore; the amount of the sample that will remain after 15 minutes is 6.25 atoms.
<u>Given:</u>
Volume of the unknown monoprotic acid (HA) = 25 ml
<u>To determine: </u>
The concentration of the acid HA
<u>Explanation:</u>
The titration reaction can be represented as-
HA + NaOH → Na⁺A⁻ + H₂O
As per stoichiometry: 1 mole of HA reacts with 1 mole of NaOH
At equivalence point-
moles of HA = moles of NaOH
For a known concentration and volume of added NaOH we have:
moles of NaOH = M(NaOH) * V(NaOH)
Thus, the concentration of the unknown 25 ml (0.025 L) of HA would be-
Molarity of HA = moles of HA/Vol of HA
Molarity of HA = M(NaOH)*V(NaOH)/0.025 L
