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Karolina [17]
4 years ago
8

Iodine-131 is a radioactive isotope. after 9.00 days, 46.0% of a sample of 131i remains. what is the half-life of 131i?

Chemistry
1 answer:
4vir4ik [10]4 years ago
6 0

For this problem, we use the integrated rate law for first order radioactive decay which is expressed as follows:<span>

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant. 

We need to calculate first the value of k from the given ration An/Ao and the time it reached that ration. We do as follows:

An = Aoe^-kt
0.46= e^-k(9)
k = 0.0863 / day

At half life, the remaining substance would be equal to one-half of the original so that An/Ao would be equal to 1/2 or 0.50. We calculate the half-life as follows:</span>

<span>
An = Aoe^-kt
0.50 = e^-0.0863(t)
<span>t = 8.03 days</span></span>

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Calculate the amount of mole of iron produced from the reaction of 15.9 grams of iron oxide.
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Answer:

0.19875

Explanation:

nFe2O3=0.099375

nFe=2nFe2O3=0.19875

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3 years ago
Diamond is a natural form of pure carbon. What number of atoms of carbon are in a 1.00-carat diamond (1.00 carat = 0.200 g)?
slega [8]

Answer:

1.004×10²²

Explanation:

The molar mass of carbon is 12 g/mol

which means that:

<u>1 mole of carbon atoms has a mass of 12 grams.</u>

Since, diamond is a allotrope of carbon.

Mass of  1.00-carat diamond in grams is:

1.00 carat = 0.200 g

<u> Since, 1 mole of C contains 6.022×10²³ atoms of C</u>

So,  

12 grams contains 6.022×10²³ atoms of C

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0.200 gram contains (6.022×10²³/ 12)×0.200 atoms of C

Thus,

<u>1 carat diamond contains 1.004×10²² atoms of C.</u>

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Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g . Solubility of NaNO3=21.9g or NaNO3 x 100 g/ 25 g =87.6. Calculate the molar mass of the dissolved compound as the sum of mass of all atoms in the molecule.

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What is the electron configuration for<br> 08<br> 16
Elis [28]

The electron configuration for Oxygen : [He] 2s²2p⁴

<h3>Further explanation</h3>

Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d, and f. The maximum number of electrons for each sub-shell is  

• s: 2 electrons  

• p: 6 electrons  

• d: 10 electrons and  

• f: 14 electrons  

Charging electrons in the sub-shell uses the following sequence:  

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  </em>

The element is Oxgen, with symbol O, and :

the atomic number=8=number of electron

the atomic mass=16

The electron configuration based on the number of electrons(for Oxygen=8), so the configuration :

\tt _8^{16}O:1s^22s^22p^4 or we can write with noble gas [He] 2s²2p⁴

3 0
3 years ago
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