Oxygen and carbon dioxide
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
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Answer:
B. CaCl + LiCO3 yields CaCO3 + LiCl is not correct
It should be CaCl2 + Li2CO3 → 2LiCl + CaCO3
Explanation:
For a reaction to be double displacement reaction there are two things we need to look for
1) There must be an interchange of the group of ions
2) The reactants must dissolve in water to release ions
A. 2RbNO3 + BeF2 yields Be(NO3)2 + 2RbF
2Rb+ + NO3- + Be^2+ + 2F- → Be(NO₃)₂ + 2RbF
This is correct
B. CaCl + LiCO3 yields CaCO3 + LiCl
This is not correct
The correct equation is:
CaCl2 + Li2CO3 → Ca2+ + 2Cl- + 2Li+ + CO3^2- → 2LiCl + CaCO3
C. Na3PO4 + 3KOH yields 3NaOH + K3PO4
3Na+ + PO4^3- + 3K+ + 3OH- → 3NaOH + K3PO4
This is correct
D. 2MgI2 + Mn(SO3)2 yields 2MgSO3 + MnI4
2Mg^2+ + 4I- + Mn^4+ + 2SO3^2- → 2 MgSO3 + MnI4
This is correct
Answer:
i think its a liquid..... im not that sure
