Answer:
Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)
If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C
Using dc = (1/4πεo)qQ/Eα we have
dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm
Note: 1meter = 10^15fentometer
Explanation:
This is well inside the atom but some eight nuclear diameters from the centre of the lead nucleus.
Can you show the question that goes with those answer pls
Answer:
26.67 mol HCl
Explanation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.
To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:
- 8.89 mol Al(OH)₃ *
= 26.67 mol HCl
Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.
Answer:
A:There must be diversity in the population.
Explanation:
Hopefully this helps!