Answer:
After 5 second 25% C-15 will remain.
Explanation:
Given data:
Half life of C-15 = 2.5 sec
Original amount = 100%
Sample remain after 5 sec = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 5 sec / 2.5 sec
Number of half lives = 2
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
Thus after 5 second 25% C-15 will remain.
Solution:
mass of the cellulose in the mixture is 0.38g
total mass of the mixture is:
3.35+0.38+8.76
=12.4g
thus the percentage of the cellulose in the mixture is:
mass of the cellulose/total mass of the mixture*100%
0.38/12.4*100%
=3%
Answer:
2 grams of salt must be added by chef to make the recipe noticeably saltier.
Explanation:
Weber's Law :
Where =
I = Initial stimulus intensity
ΔI = Difference threshold
k = Weber constant
According to Weber’s constant for saltiness , k= 
Amount of salt required for the recipe , I= 10 g
2 grams of salt must be added by chef to make the recipe noticeably saltier.
Answer:
Activation energy, in chemistry, the minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation or physical transport.
Explanation:
Both work together and the human activity is what population counts on and the activity is what effects the population
