Answer:
The correct answer is - 9935 years approximately.
Explanation:
Let z be the age in years to be found:
(15300 disintegrations) x (1.0 g / 0.250 g) / (1.84×10^4 disintegrations)
= 3.3260
half life of carbon = (1/2)^(z/5730 yr)
Solve for z
3.3260 = (1/2)^(z/5730)
Take the log of both sides:
log 3.3260 = (z/5730) log (1/2)
log 3.3260 / log (1/2) = z/5730
z = 5730 log 3.3260 / log (1/2)
= 1.73378816*5730
= 9935 years approximately.
Answer:
2.943 °C temperature change from the combustion of the glucose has been taken place.
Explanation:
Heat released on combustion of Benzoic acid; :
Enthaply of combustion of benzoic acid = 3,228 kJ/mol
Mass of benzoic acid = 0.590 g
Moles of benzoic acid = 
Energy released by 0.004831 moles of benzoic acid on combustion:

Heat capacity of the calorimeter = C
Change in temperature of the calorimeter = ΔT = 2.125°C



Heat released on combustion of Glucose: :
Enthaply of combustion of glucose= 2,780 kJ/mol.
Mass of glucose=1.400 g
Moles of glucose =
Energy released by the 0.007771 moles of calorimeter combustion:

Heat capacity of the calorimeter = C (calculated above)
Change in temperature of the calorimeter on combustion of glucose = ΔT'



2.943 °C temperature change from the combustion of the glucose has been taken place.
Answer:

Explanation:
The molecular mass of a monomer unit is:
C₂H₃Cl = 2×12.01 + 3×1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 u
For 1565 units,

I would say B because c and d would decrease competition and a would do the same, or just kill the ecosystem.
Hope this helps and don't forget to hit that heart :)