The theory of evolution was proposed by Darwin.
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:
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Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
Answer:
% purity of limestone = 96.53%
Explanation:
Question (4).
Weight of impure CaCO₃ = 25.9 g
Molecular weight of CaCO₃ = 40 + 12 + 3(16)
= 100 g per mole
We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters
From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of
CO₂.
∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g
∴ 1 liter of CO₂ will be produced by CaCO₃ = 
∴ 5.6 liters of CO₂ will be produced by CaCO₃ = 
= 25 g
Therefore, % purity of CaCO₃ = 
= 
= 96.53 %