The 2 L of sucrose stock solution would contain similar
concentration with the 100 mL aliquot. Therefore the concentration of aliquot
is still 2 M.
The molar mass of sucrose is 342.3 g / mol. Therefore the
mass in a 100 mL (0.1 L) aliquot is:
mass = (2 mol / L) * 0.1 L * (342.3 g / mol)
<span>mass = 68.46 g</span>
I don't even know what the answer is
Answer:
The artifact is 570 years old. That is, 5.7 × 10² years.
Explanation:
Radioactive decay follows first order reaction kinetics.
Let the initial activity for fresh Carbon-14 be A₀
And the activity at any other time be A
The rate of radioactive decay is given by
dA/dt = - KA
dA/A = - kdt
Integrating the left hand side from A₀ to A₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)
In [(A₀/2)/A₀] = - k t(1/2)
In (1/2) = - k t(1/2)
- In 2 = - k t(1/2)
k = (In 2)/t₍₁,₂₎
t(1/2) is given in the question to be 5.73 × 10³ years
k = (In 2)/5730 = 0.000121 /year
dA/A = - kdt
Integrating the left hand side from A₀ to A and the right hand side from 0 to t
In (A/A₀) = - kt
A/A₀ = e⁻ᵏᵗ
A = A₀ e⁻ᵏᵗ
A = 2.8 × 10³ Bq.
A₀ = 3.0 × 10³ Bq.
2.8 × 10³ = 3.0 × 10³ e⁻ᵏᵗ
0.9333 = e⁻ᵏᵗ
e⁻ᵏᵗ = 0.9333
-kt = In 0.9333
- kt = - 0.06899
t = 0.06899/0.000121 = 570.2 years = 5.7 × 10² years
Answer:
C5H12.
Explanation:
The compound will be C5H12.
Now consider the general formula for the combustion of an alkane;
CnH2n+2 + 3n+1/2O2-----> nCO2 + (n+1) H2O
If we substitute n=5, we have;
C5H12 + 8O2-----> 5CO2 + 6H2O
Hence the answer above.
Answer: B) The identity of the solvent
Explanation:
Basically, the solvent is the liquid in which a solute is dissolved in. But the solute is the material to be dissolved.
Now in this case, the solute in the first solution is glucose and the solute in the second solution is an unidentified covalent solid material.
This means that:
• the identity of the solute cannot be identical in each solutions, which also means that the freezing points and densities of the solutions cannot be identical too.
• the only thing that is sure to be identical in the solution is the identity of the solvent.
