A magical or medicinal potion/solution
The half life of carbon-14 is 5700 years. (Half life is the time taken by a radioactive isotope to decay by half of its original mass).
Let A₀ be the initial amount of carbon-14 that is found in living matter (t=0 years), to determine when there was 44.5% of A₀ left.
44.5 = 100 × (1/2)^n, where n is the number of half lives
0.5^n = 0.445
n = log 0.445/log 0.5
n = 1.168
But 1 half life is 5700 years
Therefore, the number of years will be 5700 × 1.168 = 6658.299725 years
≈ 6658.30 years
Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
Answer:DNA
Explanation:I already do the test
