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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<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
scandium is a transition metal and shows a valency of 3.
Well you would think yeah because it’s a liquid but the answer is no
The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.
Further Explanation:
To solve this problem, follow the steps below:
- Write the balanced chemical equation for the given reaction.
- Convert the mass of calcium carbonate into moles.
- Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
- Convert the number of moles of calcium oxide into mass.
Solving the given problem using the steps above:
STEP 1: The balanced chemical equation for the given reaction is:
STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.
STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.
For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,
STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.
Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.
Therefore,
Learn More
- Learn more about stoichiometry brainly.com/question/12979299
- Learn more about mole conversion brainly.com/question/12972204
- Learn more about limiting reactants brainly.com/question/12979491
Keywords: thermal decomposition, stoichiometry
