The age of painting was determined from the decay kinetics of the radioactive Carbon -14 present in the painting sample.
Given that the half life of Carbon-14 is 5730 years.
Radioactive decay reactions follow first order rate kinetics.
Calculating the decay constant from half life:
λ
= 
= 
Setting up the radioactive rate equation:
Where 
k = decay constant =
ln 0.125 = 
-2.079=
t=
= 17185 years
t = 17185 years
Therefore age of the painting based in the radiocarbon -14 dating studies is 17185 years
Answer is: 6.022·10²² molecules of glucose.
c(glucose) = 100 mM.
c(glucose) = 100 · 10⁻³ mol/L.
c(glucose) = 0.1 mol/L; concentration of glucose solution.
V(glucose) = 1 L; volume of glucose solution.
n(glucose) = c(glucose) · V(glucose).
n(glucose) = 0.1 mol/L · 1 L.
n(glucose) = 0.1 mol; amount of substance.
N(glucose) = n(glucose) · Na (Avogadro constant).
N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.
N(glucose) = 6.022·10²².
We are given that the balanced chemical reaction is:
cacl2⋅2h2o(aq) +
k2c2o4⋅h2o(aq) --->
cac2o4⋅h2o(s) +
2kcl(aq) + 2h2o(l)
We known that
the product was oven dried, therefore the mass of 0.333 g pertains only to that
of the substance cac2o4⋅h2o(s). So what we will do first is to convert this
into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is
molar mass of cac2o4 plus the
molar mass of h2o.
molar mass cac2o4⋅h2o(s) = 128.10
+ 18 = 146.10 g /mole
moles cac2o4⋅h2o(s) =
0.333 / 146.10 = 2.28 x 10^-3 moles
Looking at
the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is
1:1, therefore:
moles k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles
Converting
this to mass:
mass k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles (184.24 g /mol) = 0.419931006 g
Therefore:
The mass of k2c2o4⋅<span>h2o(aq) in
the salt mixture is about 0.420 g</span>
