Half life is the time taken by a radioactive isotope to decay by half its original mass.
The original mass is 200 g
Time taken is 60 hours
Final mass is 25 g
Therefore;
Final mass = Original mass × (1/2)^n; where n is the number of half lives.
25 = 200 (1/2)^n
1/8 = (1/2)^n
n = 3
Three half lives = 60 hours
1 half lives = 20 hours
Therefore; the half life of the radioactive nucleus is 20 hours
Explanation:
The given data is as follows.
Solvent 1 = benzene, Solvent 2 = water
= 2.7, 
= 100 mL
= 10 mL, weight of compound = 1 g
Extract = 3
Therefore, calculate the fraction remaining as follows.
![f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5B1%20%2B%20K_%7Bp%7D%28%5Cfrac%7BV_%7BS_%7B2%7D%7D%7D%7BV_%7BS_%7B1%7D%7D%7D%29%5D%5E%7B-n%7D)
= ![[1 + 2.7(\frac{100}{10})]^{-3}](https://tex.z-dn.net/?f=%5B1%20%2B%202.7%28%5Cfrac%7B100%7D%7B10%7D%29%5D%5E%7B-3%7D)
= 
= 
Hence, weight of compound to be extracted = weight of compound - fraction remaining
= 1 -
= 0.00001
or, = 
Thus, we can conclude that weight of compound that could be extracted is .
Answer:
N₂+3H₂ ⇄2NH₃ is a thermochemical reaction whereas A+BC⇄AB is not.
A+BC⇄AB is a reaction of pure a element with a compound while N₂+3H₂ ⇄2NH₃ is a reaction between two pure elements.
Explanation:
Let A+BC⇄AB be equation i and N₂+3H₂ ⇄2NH₃ be equation ii.
The two reactions differ in that ii is a thermo-chemical reaction whereas i is not. This is because energy is included in reaction ii but not included in reaction i.
Also i is a reaction of pure a element with a compound while ii is a reaction between two pure elements. The compound is BC while the pure element is A.
One thing that does not change is the chemical composition of water, which is still H2O. And maybe mass, if all of the particles remain inside the beaker, which was never mentioned in the question so I am not sure.
Answer:
3.8 x 10²⁴molecules
Explanation:
Given parameters:
Number of moles = 6.32moles
Unknown:
Number of molecules = ?
Solution:
The number of moles can be used to derive the number of molecules found within a substance.
Now,
1 mole of substance contains 6.02 x 10²³ molecules
6.32 mole of PBr₃ will contain 6.32 x 6.02 x 10²³ = 3.8 x 10²⁴molecules
