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.
Answer:
4057.85 g/mol
Explanation:
Hello, the numerical procedure is shown in the attached file.
- In this case, since we don't have the density of the protein, we must assume that the volume of the solution is solely given by the benzene's volume, in order to obtain the moles of the solute (protein).
-Van't Hoff factor is assumed to be one.
Best regards.
Answer:
A) (3.2g)
Explanation:
Did you reposed this? Because I remember answering this
Answer:
Explanation:
Let the number of half lives be x
<u>Solve this equation to find the value of x:</u>
- 125*(1/2)ˣ = 3.90625
- (0.5)ˣ = 3.90625 / 125
- (0.5)ˣ = 0.03125
- log (0.5)ˣ = log 0.03125
- x = log 0.03125 / log 0.5
- x = 5
