Answer:
After 6sec 5 atoms will be left.
Explanation:
Given data:
Total atoms of Highlinium = 40
Half life = 2 sec
Atoms left after 6sec = ?
Solution:
First of all we will calculate the number of half lives passes during 6sec.
Number of half lives = Time elapsed/ half life
Number of half lives = 6 sec / 2sec
Number of half lives = 3
Now we will calculate the number of atoms left
At time zero = 40 atoms
At 1st half life = 40 atoms/2 = 20 atoms
At 2nd half life = 20 atoms/2 = 10 atoms
At 3rd half life = 10 atoms / 2 = 5 atoms
Thus, after 6sec 5 atoms will be left.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
c=9,50*10⁻² mol/L
Ammonia solution is a weak electrolyte.
The dissociation constant of ammonium hydroxide is:
K=1.78*10⁻⁵
[OH⁻]=√(Kc)
pH=14-pOH=14+lg[OH]⁻
pH=14+lg(√(Kc))
pH=14+lg(√(1.78*10⁻⁵*9.50*10⁻²))=8.23
pH=8.23
The first is focus the second is fault the third is epicenter
Answer:
star 3
Explanation:
the color determines the fusion in between