First, we write the reaction equation:
2KI + PbNO₃ → K₂NO₃ + PbI₂
The molar ratio of KI to PbNO₃ is 2 : 1
Moles of PbNO₃ present:
Moles = concentration (M) x volume (dm³)
= 0.194 x 0.195
= 0.038
Moles of KI required = 2 x 0.038 = 0.076 moles
concentration = moles / volume
volume = moles / concentration
= 0.076 / 0.2
= 0.38 L = 380 ml
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.
Only one of the listed choices are correct here:
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