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skad [1K]
2 years ago
15

Thallium-201 has a half-life 73 hours. If 4.0 mg of thallium-201 disintegrates over a period of 5.0 °, how many milligrams of th

allium-201 will remain?
Chemistry
1 answer:
Anika [276]2 years ago
5 0

<u>Given:</u>

Half life of Thallium-201 (T1/2)= 73 hrs

Initial amount of thallium (A₀) = 4.0 mg

Disintegration Time period (t) = 5.0 hrs

<u>To determine:</u>

Amount of thallium remaining, A

<u>Explanation:</u>

The radioactive half-life (T1/2) can be related to the amounts of the radio active element through the relation:

A = A_{0} e^{-0.693t/T1/2}

A = 4.0 * e^{-0.693*5/73} = 3.815 mg

Ans: Amount of thallium-201 remaining is 3.815 mg

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Read 2 more answers
4Cr(s)+3O2(g)→2Cr2O3(s) calculate how many grams of the product form when 21.4 g of O2 completely reacts
weqwewe [10]

Answer:

= 67.79 g

Explanation:

The equation for the reaction is;

4Cr(s)+3O2(g)→2Cr2O3(s)

The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;

moles O2 = 21.4 g / 32 g/mol

                =0.669 moles

Using mole ratio, we get the moles of Cr2O3;

moles Cr2O3 = 0.669 x 2/3

                       =0.446 moles

but molar mass of Cr2O3 is 151.99 g/mol

Hence,

The mass Cr2O3 = 0.446 mol x 151.99 g/mol

                            <u> = 67.79 g </u>

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