Answer:
A microorganism, or microbe, is a microscopic organism, which may exist in its single-celled form or a colony of cells.
Explanation:
Explanation:
Moles of phosphorus pentachloride present initially = 2.5 mol
Moles of phosphorus trichloride at equilibrium = 0.338 mol

Initially
2.5 mol 0 0
At equilibrium:
(2.5 - x) mol x x
So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol
Mass of 0.338 moles of phosphorus trichloride at equilibrium:
= 0.338 mol × 137.5 g/mol = 46.475 g
Moles of phosphorus pentachloride present at equilibrium :
= (2.5 - 0.338) mol = 2.162 mol
Mass of 2.162 moles of phosphorus pentachloride at equilibrium:
= 2.162 mol × 208.5 g/mol = 450.777 g
Moles of chloride gas present at equilibrium : 0.338 mol
Mass of 0.338 moles of chloride gas at equilibrium:
= 0.338 mol × 71 g/mol = 23.998 g
<span>For equation A + 3B + 2C ---> 2D,
1 mole of A will produce 2 moles of D
3 moles of B will produce 2 moles of D, so 1 mole of B will produce 2/3 moles of D
2 moles of C will produce 2 moles of D, so 1 mole of C will produce 1 mole of D
If only 1 mole of B is present, only 2/3 moles of D can be produced. This is regardless of the number of moles of A and C. B is the limiting reactant and the maximum number of moles of D expected is 2/3.</span>
Answer:
The age of the sample is 4224 years.
Explanation:
Let the age of the sample be t years old.
Initial mass percentage of carbon-14 in an artifact = 100%
Initial mass of carbon-14 in an artifact = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final mass percentage of carbon-14 in an artifact t years = 60%
Final mass of carbon-14 in an artifact = ![[A]=0.06[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D0.06%5BA_o%5D)
Half life of the carbon-14 = 

![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D%5Ctimes%20t%7D)
![0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}](https://tex.z-dn.net/?f=0.60%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7B5730%20year%7D%5Ctimes%20t%7D)
Solving for t:
t = 4223.71 years ≈ 4224 years
The age of the sample is 4224 years.