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: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of 
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.0787 moles of 
require=
of
Thus is the limiting reagent as it limits the formation of product and acts as the excess reagent. (10.0-0.0787)= 9.92 moles of are left unreacted.
Mass of 
Thus 19.9 g of remains unreacted.
Answer: 317 joules
Explanation:
The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
In this case,
Q = ?
Mass of aluminium = 50.32g
C = 0.90J/g°C
Φ = (Final temperature - Initial temperature)
= 16°C - 9°C = 7°C
Then, Q = MCΦ
Q = 50.32g x 0.90J/g°C x 7°C
Q = 317 joules
Thus, 317 joules of heat is gained.
Answer:
1.69.
Explanation:
- The solution = 12.0 / 7.11 = 1.687 = 1.69.
- The rule of significant figures for division states that: the results are reported to the fewest significant figures.
- 12.0 contains 3 significant figures.
- 7.11 contains 3 significant figures.
So, the solution should contain 3 significant figures.
- Now, the issue id of rounding; In a series of calculations, carry the extra digits through to the final result, then round.
- If the digit to be removed is equal to or greater than 5, the preceding digit is increased by 1.
- The digit that should be removed is 7 that is larger than 5 so increase the preceding digit by 1.
