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.
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Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
Answer:
1. 176 × 10^12 W ; 78400000000
Explanation:
Given the following :
Fall rate = 2,400,000kg/s
Average height of fall = 50m
Gravitational Potential of falling water = mgh = mass × acceleration due to gravity × height =
How many 15 W LED light bulbs could it power?
Recall : power = workdone / time
Workdone = gravitational potential energy
Mass of water = density * volume
Density of water = 1 * 10^3kg/m^3
Rate of fow = volume / time = 2400000
Hence,
Power = 1000 * 2,400,000 * 9.8 * 50
Power = 1176000000000
Power = 1. 176 × 10^12 W
How many 15 W LED light bulbs could it power?
1176000000000 / 15 = 78400000000
= 78400000000 15 W bulbs
B. The answer is: All nucleotides have a phosphorus atom that can be replaced with 32P.
Nucleotides contain a nitrogenous base, a five-carbon sugar, and, at least, one phosphate group. Exactly that phosphate group in the nucleotide has the phosphorus atom. Therefore, the phosphorus atom in the nucleotide can be replaced with radioactive phosphorus-32 (32P).
Carboxylic acid...........