CHAPTER 19 REGULAR HOMEWORK:
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If the protons change the element changes so chlorine will always have 17 protons.. never more never less because then it would’nt be chlorine it would be one of its periodic table neighbors
Take the mass number 35 an subtract the number of protons 17
Z is the atomic/proton number
A is the mass number
Caesium will always have 55 protons
(Input symbol, remember neon is Ne NOT ne.)
So find the element on the periodic table with 33 as their atomic number aka Arsenic
28 + 3 = 31 (proton)
Does the above isotope lie within the band of nuclear stability? (Compare number of neutrons to number of protons with what is given in graph.)
Does the above isotope lie within the band of nuclear stability?
Identify the missing particles by balancing the mass and atomic numbers in each of the following nuclear decay equations.
Does Ga-67, the above isotope, lie within the band of nuclear stability?
How many α and β particles are emitted in the decay series that begins with Np-237 and finally produces the stable isotope Bi-209?
First, how many α (alpha) emissions?
ln (Rt/Ro) = -kt
ln(546/927) = -k5
t1/2= 0.693/k = 6.56
1238 = R(2.64e10)
t1/2=0.693/4.7e8=14744680.9 min * 1 yr/525600 min = 28.12 yrs
ln(954/1238) = -4.7e-8(t)
t=5544442.166 min * 1hr/60 min * 1 day/24 hr *1 yr/ 365 day=10.5
t1/2 = 5730 yrs * 525600 min/year = 3011688000 min = 0.693/k
25.7/3.7 = 6.954 dis/min*g
3433455027*1 yr/525600 min=6532.33yrs
Hint, (want the number of C-14 atoms in 1 g)/(total C atoms in 1 g)x100%. Use rate = kN for number of C-14 and what you know about moles and Avogadro’s number for the total number of C atoms in 1 g.
1g * 1mol/12.011g * 6.022e23/1 mol = 5.01e22
%= 6.65e10/5.01e22 * 100 = 1.33e-10
# of protons = 15
15 * 1.007825 = 15.117
# of neutrons=15
15* 1.008665= 15.129975
mass defect = 30.24755-29.9783
0.26905 * 931.5 MeV/1 * 250.62 MeV
250.62MeV/30 = 8.354
When 239Pu is used in a nuclear reactor, one of the fission events that occurs is
The atomic mass of each atom is given below its symbol in the equation. Find the energy released (in kJ) when 5.10 g of plutonium undergoes this particular fission. Hint given in feedback.
Hint, (1) Calculate the change in mass, expressed in atomic mass units, that accompanies this fission event. (2) Convert the mass loss to joules per fission. (3) Find the energy released per mole of plutonium. . .
=0.193 * 1.66054e-27kg/1 = 3.2e-28 kg
2.88e-11J/atoms * 5.1 g Pu * 1mol/239 g * 6.022e23/1mol=3.7e11 J
t1/2= 0.693/k = 28.1 yr
A= Ao(1/2)^n= 10(1/2)^3.523
1 kg U * 1000 g U / 11 g U * 1 mol/235.05 g * 1.9e10kg/1mol U * 1 kg coal/2.8e4 kJ=2.89e6 kg
Remember to use the relation between the mass of S in coal and the mass of SO2.
2.89e6 kg * 0.76/100 * 100 kg SO2/50.1 kg S = 4.38e4
Remember that the rate of effusion is proportional to the average velocity which is inversely proportional to the square root of the mass. (vrms> = (3RT/M)0.5)
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