1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Viefleur [7K]
3 years ago
14

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

the mass (in mg) remaining after t years. Then we know the following. y ( t ) = y ( 0 ) e k t = ⋅ e k t Since the half-life is 30 years, then y ( 30 ) = mg. Thus, e 30 k = . Therefore, k = . Now, remembering that ln x n = n ln x and that e ln z = z , we have y ( t ) = 180 e ( t / 30 ) ( − ln 2 ) = mg. How much of the sample remains after 70 years? After 70 years we have the following. y ( 70 ) = 180 ⋅ 2 = mg (Round your answer to two decimal places.) After how long will only 1 mg remain? To find the time at which only 1 mg remains, we must solve 1 = y ( t ) = 180 ( 2 − t / 30 ) , and so we get the following. t = − 30 log 2 ( ) Hence, we conclude the following. t = yr (Round your final answer to one decimal place.)
Chemistry
1 answer:
Stels [109]3 years ago
7 0

Explanation:

1) Initial mass of the Cesium-137=N_o= 180 mg

Mass of Cesium after time t = N

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

Half life of the cesium-137 = t_{1/2}=30 years[p/tex]where,
[tex]N_o = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

t_{\frac{1}{2}} = half life of the isotope

\lambda = rate constant

N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}

Now put all the given values in this formula, we get

N=180mg\times e^{-\frac{0.693}{30 years}\times t}

Mass that remains after t years.

N=180 mg\times e^{0.0231 year^{-1}\times t}

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

N_o=180 mg

t_{1/2}= 30 yeras

N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

N_o=180 mg

t_{1/2}= 30 yeras

N = 1 mg

t = ?

1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}

\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

You might be interested in
Formula of a substance in water is followed by what symbol
snow_lady [41]
The substance is followed by H2O
8 0
3 years ago
tate whether the following changes are physical or chemical for rancidipication fixation of water 2 tearing of paper 3 rusting o
damaskus [11]

Answer: Physical change : tearing of paper, fixing of wtaer

Chemical change:  rusting of iron ,  electrolysis of water​, Rancidification

Explanation:

Physical change is a change in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.

Example:  tearing of paper, fixing of wtaer

Chemical change is a change in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.

Example: rusting of iron ,  electrolysis of water​, Rancidification

3 0
2 years ago
Determine all values of hh and k for which the system has no solution
solniwko [45]
Explain more so i could answer it!!
6 0
2 years ago
An atom of uranium-238 undergoes radioactive decay to form an atom of thorium-234. Which type of nuclear decay has occurred?
WINSTONCH [101]

Answer:-

Alpha decay

Explanation:-

Uranium 238 has atomic number 92 and mass number 238.

Thorium 234 has atomic number 90 and mass number 234.

So, the change in atomic number as uranium 238 disintegrates into thorium234 = 92 – 90 = 2

So, the change in mass number as uranium 238 disintegrates into thorium234= 238 – 234 = 4

We know that when an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2.

So when an atom of uranium 238 undergoes radioactive decay to form an atom of thorium-234, alpha decay has occurred.

5 0
2 years ago
7. What is the molarity, M, of the dilution solution of KBr, given:
sineoko [7]

Answer:

Hello There!!

Explanation:

I believe the answer is c. Volume of the dilute solution is 100.00 mL.

hope this helps,have a great day!!

~Pinky~

7 0
2 years ago
Read 2 more answers
Other questions:
  • What are weak bonds that allow flexibility in an enzyme
    9·1 answer
  • 3. Find the mass of 4.77 x 1022 atoms of scandium<br> (Sc).
    5·1 answer
  • Phenolphthalein indicator was added, and the solution in the flask was titrated with 0.215M NaOH until the indicator just turned
    13·2 answers
  • What mixture can be separated by using a filter?
    9·1 answer
  • 5. How many atoms are in 1.00 mol of carbon atoms?
    12·1 answer
  • Limiting Reactant problem
    8·1 answer
  • How many moles of NaCl are contained in 144 mL of a 1.65 M NaCl solution?
    9·1 answer
  • When heated, magnesium combines readily with excess oxygen in the air to produce magnesium oxide, as shown in the following unba
    12·1 answer
  • En la formación del compuesto iónico entre litio y oxígeno. ¿Cuántos electrones recibe oxígeno del átomo de litio?
    14·1 answer
  • What is the correct definition of Earth's energy budget?
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!