### CHAPTER 19 REGULAR HOMEWORK:

### NUCLEAR CHEMISTRY

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**Question 1**

**Explanation**

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

**Question 2**

**Explanation**

Take the mass number 35 an subtract the number of protons 17

35-17=18

**Question 3**

**Explanation**

Z is the atomic/proton number

**Question 4**

**Explanation**

A is the mass number

**Question 5**

**Explanation**

Caesium will always have 55 protons

**Question 6**

**Explanation**

133-55=78

**Question 7**

**Question 8**

**Question 9**

(Input symbol, remember neon is Ne NOT ne.)

**Explanation**

Z=proton number

So find the element on the periodic table with 33 as their atomic number aka Arsenic

**Question 10**

**Explanation**

28 + 3 = 31 (proton)

aka Gallium

**Question 11**

**Explanation**

8:9

**Question 12**

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.)

**Question 13**

**Explanation**

29:39

**Question 14**

Does the above isotope lie within the band of nuclear stability?

**Question 15**

**Explanation**

94:145

1.542

**Question 16**

Identify the missing particles by balancing the mass and atomic numbers in each of the following nuclear decay equations.

**Question 17**

Does Ga-67, the above isotope, lie within the band of nuclear stability?

**Question 18**

**Question 19**

**Question 20**

**Question 21**

**Question 22**

**Question 23**

^{117}Sb (For one type of decay more than one choice is correct.)

**Question 24**

^{83}Se

**Question 25**

^{42}Ar

**Question 26**

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?

**Question 27**

**Question 28**

**Explanation**

ln (Rt/Ro) = -kt

ln(546/927) = -k5

k=0.106

t1/2= 0.693/k = 6.56

**Question 29**

^{10}atoms of Sr-90 as the only radioactive element. The absolute disintegration rate of this sample is measured as 1238 disintegrations per minute. Calculate the half-life (in years) of Sr-90.

**Explanation**

R=KN

1238 = R(2.64e10)

k=4.689e-8

t1/2=0.693/4.7e8=14744680.9 min * 1 yr/525600 min = 28.12 yrs

**Question 30**

**Explanation**

ln(954/1238) = -4.7e-8(t)

t=5544442.166 min * 1hr/60 min * 1 day/24 hr *1 yr/ 365 day=10.5

**Question 31**

**Explanation**

t1/2 = 5730 yrs * 525600 min/year = 3011688000 min = 0.693/k

k=2.3e-10

ln(Rt/Ro)=-kt

25.7/3.7 = 6.954 dis/min*g

ln(6.945/15.3)=-(2.3e-10)t

3433455027*1 yr/525600 min=6532.33yrs

**Question 32**

**Explanation**

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.

R=kN

15.3=2.3e-10(N)

N=6.65e10

1g * 1mol/12.011g * 6.022e23/1 mol = 5.01e22

%= 6.65e10/5.01e22 * 100 = 1.33e-10

**Question 33**

**Explanation**

# of protons = 15

15 * 1.007825 = 15.117

# of neutrons=15

15* 1.008665= 15.129975

mass defect = 30.24755-29.9783

=0.26905

0.26905 * 931.5 MeV/1 * 250.62 MeV

250.62MeV/30 = 8.354

**Question 34**

**Question 35**

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.

**Explanation**

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. . .

Mass reactants-products

(1.008665+239.052)-(95.916+149.917+1.008665(4))

=240.060665-239.9

=0.193 * 1.66054e-27kg/1 = 3.2e-28 kg

E=mc^2

=(3.2e-28)(3e8)^2

=2.88e-11

2.88e-11J/atoms * 5.1 g Pu * 1mol/239 g * 6.022e23/1mol=3.7e11 J

=3.7e8 J

**Question 36**

**Explanation**

t1/2= 0.693/k = 28.1 yr

k=0.024662

A= Ao(1/2)^n= 10(1/2)^3.523

n=(2044-1945)/28.1=3.523

A=.869

**Question 37**

^{4}kJ of energy when it is burned. Fission of one mole of U-235 releases 1.9 x 10

^{10}kJ. Calculate the number of kg of coal needed to produce the same amount of energy as the fission of 1 kg of U-235 (235.05g/mol).

**Explanation**

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

**Question 38**

**Explanation**

Remember to use the relation between the mass of S in coal and the mass of SO_{2}.

2.89e6 kg * 0.76/100 * 100 kg SO2/50.1 kg S = 4.38e4

**Question 39**

_{6}. The atomic masses of the two principal isotopes of uranium are 235.0493 and 238.0508 for U-235 and U-238, respectively, and that of F is 18.9984. Use 4 decimal places.

**Explanation**

Remember that the rate of effusion is proportional to the average velocity which is inversely proportional to the square root of the mass. (v_{rms>} = (3RT/M)^{0.5})

**Question 40**

**Explanation**

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