Answer:
17202.6 years
Explanation:
Activity of the living sample (Ao) = 160 counts per minute
Activity of the wood sample (A) = 20 counts per minute
Half life of carbon-14 = 5730 years
t= age of the artifact
From;
0.693/t1/2= 2.303/t log Ao/A
Then;
0.693/ 5730= 2.303/t log Ao/A
Substituting values;
0.693/5730= 2.303/t log (160/20)
Then we obtain;
1.209×10^-4 = 2.0798/t
t= 2.0798/1.209×10^-4
Thus;
t= 17202.6 years
Therefore the artifact is 17202.6 years old.
Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use <span>Henderson–Hasselbalch equation:
pH = pKa + log(c(</span>NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
<span>[H</span>₃O⁺] = 10∧(-4,02).
<span>[H</span>₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.
Answer:
They include frequency, period,speed,amplitude and phase
