100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Chem 162 Practice Final Exam KEY. $10.99   Add to cart

Exam (elaborations)

Chem 162 Practice Final Exam KEY.

 1 view  0 purchase

This is a comprehensive and detailed Practice Final Exam KEY for Chem 162. *Essential Study Material!!

Preview 2 out of 5  pages

  • September 13, 2024
  • 5
  • 2020/2021
  • Exam (elaborations)
  • Questions & answers
All documents for this subject (4)
avatar-seller
anyiamgeorge19
University of Hawaii at Manoa Chemistry 162 Professor Gary
ANSWER KEY

Final Practice Exam

*NOTE: The first 14 problems are specifically related to Chapter 20
material. Additionally, be aware of the constants & equations for
Chapter 20 given on Exam 3 Constants Sheet. These are the EXACT
constants & equations sheets you will be given on the final exam!*
1. Is an isotope and an allotropy the same thing?
a) yes b) no c) sometimes d) not enough info e) none of the above

2. Complete the following nuclear reaction and pick the correct product; 220Rn  4He + ____
a) 224Ra b) 216At c) 216Po d) 216Rn e) none 216
84 Po



3. Which is the correct symbol for a bismuth-211 atom that has decayed by beta emission?
a) 211Pb b) 212Bi c) 212Pb d) 210Tl e) None 211Bi  01 e + 211Po

4. What percentage of cesium chloride made from cesium-137 (t1/2= 30 yr; beta emitter) remains
after 150 yr? a) 31% b) 13% c) 9.1% d) 0.13 % e) 3.1 %
0.693
The chemical product is BaCl2. Recall that for a first order process k =
t1 2
[A] 0
So k = 0.693/30 yr = 2.30 × 10–2/yr. Also, ln  kt  [A]t  [A]0 exp(  kt) 
[A] t
[A] 0  [A] 0 
 exp[ (2.30  10 2 /yr)(150 yr)]   3.13  10 2
[A] t [A] t
so 3.1% of the original sample remains. Or that 5 half life x 30yrs = 150 yrs 1/25 = 1/32
=0.03125

5. Although lead-164 has two magic numbers, 82 protons and 82 neutrons, this isotope is unknown.
Lead-208, however, is known and stable. Which statement best accounts for the nonexistence of lead-
164?
a) not enough neutrons
b) too many protons
c) Atomic number is more than bismuth
d) too many neutrons
e) neutron-proton ratio is to high
The neutron-to-proton ratio of lead–164 is too low. It has too few neutrons for the number of
protons in the nucleus.

6. Give the composition of each of the following. a) alpha particle b) beta particle c) positron
(a) The alpha particle is composed of helium nuclei, and has a 2+ charge.
(b) Beta particles are electrons.
(c) Positrons are particles with a +1 charge, and have the same mass as the electron.
(d) A deuteron consists of one neutron and one proton, i.e. the nucleus of a deuterium atom.

, University of Hawaii at Manoa Chemistry 162 Professor Gary

7. Radionuclides of high atomic number are far more likely to be alpha emitters than those of low
atomic number. Offer an explanation for this phenomenon.
The loss of an alpha particle is the most effective way to move the nuclide toward the band of
stability, if the unstable nuclide lies to the right and above the band of stability, because the
alpha particle has a relatively large mass

8. Write a balanced nuclear equation for each of the following changes.
242 4 238
a) alpha emission from plutonium-242 (a) 94 Pu  2 He + 92 U
28 0 28
b) beta emission from magnesium-28 (b) 12 Mg  1 e + 13 Al


9. If we begin with 3.00 mg of iodine-131 (t1/2 - 8.07 days), how much remains after 6 half-life
periods? Six half–life periods correspond to the fraction 1/64 of the initial material. That is,
one sixty–fourth of the initial material is left after 6 half lives: 3.00 mg × 1/64 = 0.0469 mg
remaining.

10. Neutron bombardment of cadmium-115 results in neutron capture (similar to electron capture
but capturing a neutron) and the release of gamma radiation. Write the nuclear equation:
48 Cd + 
115 1 116
48 Cd + 0 n 


11. A sample of waste has a radioactivity, caused solely by strontium-90 (beta emitter, t1/2 = 28.1
yr), of 0.245 Ci g-1. How many years will it take for its activity to decrease to 1.00 × 10-6 Ci g-1?
For a first–order process, the rate constant is related to the half–life by the equation:
k = 0.693/t1/2 Hence, k = 0.693/28.1 yr = 2.47 × 10–2 yr–1
Also, for a first–order process, the concentration varies with time according to the equation:
A  0
ln = kt
A  t
which allows us to solve for the time, t, for the activity to decrease by the specified amount:

t=
1
 ln
 A 0
=
1
 ln

0.245 Ci g = 502 yr
k   t 2.47  10 yr
A 2 1
 
1.00  10 6 Cig 
12. Which of the following nuclides are most likely to decay via beta decay?
a) I-131 b) Ar-40 c) F-18 d) Zr-90 e) Pb-206

13. Write a nuclear equation to describe the spontaneous fission of Am to form I-134 and Mo-

107. Determine how many neutrons are produced in the reaction.
a) 0 b) 1 c) 2 d) 3 e) 4

14. Determine the binding energy of an O-16 nucleus. The O-16 nucleus has a mass of 15.9905
amu. A proton has a mass of 1.00728 amu, a neutron has a mass of 1.008665 amu, and 1 amu is
equivalent to 931 MeV of energy.
a) 8.84 MeV
b) 128 MeV
c) 138 MeV
d) 78.1 MeV
e) 38.2 MeV

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller anyiamgeorge19. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $10.99. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

79650 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$10.99
  • (0)
  Add to cart