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leva [86]
1 year ago
8

A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod

ine-131 are now present?
Chemistry
1 answer:
sertanlavr [38]1 year ago
7 0

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

brainly.com/question/27891343

#SPJ4

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What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m
r-ruslan [8.4K]

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2


Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol                               2 mol
0.100 mol                           0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.



4 0
3 years ago
What is the percent of Cr in Cr2O3
IgorLugansk [536]
The percentage of Chromium in Chromium Oxide is calculated as follow,

Step 1:
           Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
     2(51.99) + 3(16) = 103.98 + 48 = 151.98 u

Step 2:
          Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
                   = \frac{103.98}{151.98} × 100
                   
                   = 68.41 %
So, the %age composition of chromium in chromium oxide is 68.41 %.
5 0
3 years ago
Read 2 more answers
What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a
Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})

Where:

Q: is the energy transferred = 5.0 MJ

k_{B}: is the Boltzmann constant = 1.38x10⁻²³ J/K  

T_{i}: is the initial temperature = 1000 K

T_{f}: is the final temperature = 500 K

Hence, the entropy change is:

\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

7 0
3 years ago
1) Describe the characteristics of molecular compounds that affect their particular physical
Nitella [24]

The types of intermolecular forces that occur in a substance will affect its physical properties, such as its phase, melting point and boiling point.
8 0
3 years ago
The energy released from 1 gram of uranium is more than 1 million times greater than the energy released from 3 grams of coal. T
SVEN [57.7K]

The energy released from 1 gram of uranium is more than 1 million times greater than the energy released from 3 grams of coal is True.

<u>Explanation:</u>

Nuclear Fission is the process in which splitting of a nucleus takes place that releases free neutrons and lighter nuclei. The fission of heavy elements like "Uranium is highly exothermic" and releases "200 million eV" compared to the energy that is released by burning coal which gives a few eV.

In the given example, it is obvious that the energy released from 1 gram of uranium is more than that of the energy released from 3 grams of coal because the amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only \frac{1}{10}^{th} part of the original nuclei is converted to energy.

5 0
3 years ago
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