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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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Give an example of competition in an ecosystem
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hbzjaj

Explanation:

competition for food

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3 years ago
What are the functions of arteries, veins, and capillaries?
lilavasa [31]

Answer:

they all have one common function which is to pump blood and Arteries carry blood away from the heart; the main artery is the aorta. ... Capillaries carry blood away from the body and exchange nutrients, waste, and oxygen with tissues at the cellular level. Veins are blood vessels that bring blood back to the heart and drain blood from organs and limbs.

Explanation:

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3 years ago
"how many moles of calcium atoms are in 2.5 moles of calcium carbonate caco3"
Montano1993 [528]

Answer: -

2.5 moles of calcium atoms are in 2.5 moles of calcium carbonate CaCO₃

Explanation: -

In order to solve such types of problems, the first step would be to write the chemical formula of the compound.

The chemical formula of calcium carbonate = CaCO₃

The chemical symbol of Calcium is Ca.

From the formula of calcium carbonate we can see that

1 mole of CaCO₃ has 1 mole of Ca

2.5 mole of CaCO₃ has \frac{1 mole Ca }{1 mole CaCO₃} x 2.5 mole CaCO₃

= 2.5 mol of Ca.

∴2.5 moles of calcium atoms are in 2.5 moles of calcium carbonate CaCO₃

5 0
3 years ago
Shelby measured the volume of a cylinder and determined it to be 54.5 cm3 . The teacher told her that she was 4.25% too high in
balu736 [363]

Answer:

V = 56.816 cm³

Explanation:

Given that,

The measured value of the cylinder is 54.5 cm³

The percent error in the measurement of the volume is 4.25%

We need to find the actual volume of the cylinder.

Firstly, we can find the error in the calculation as follows :

E=54.5 \times \dfrac{4.25}{100}=\pm2.316

It is mentioned in the problem that, the teacher told her that she was 4.25% too high in her determination of the volume. It would mean that,

Total volume = 54.5 + 2.316 = 56.816 cm³

Hence, the actual volume of the cylinder is 56.816 cm³.

7 0
2 years ago
Calculate the concentration of an anthracene solution which produced a fluorescence intensity ( I ) of 775 when the irradiance o
Kay [80]

Answer:

The concentration of an anthracene solution is  c = 3.560 *10^{-5}M

Explanation:

From the question we are told that

       The incident beam P_o is =1532

         The fluorescence intensity is  I = 775

          The length of the medium is b = 0.875 cm

         The molar extinction coefficient is  \epsilon = 9.5 *10^3 M^{-1} cm^{-1}

          The proportionality  constant k = 0.30

 

According to Lambert law the Absorbance of the anthracene solution is mathematically represented as

              A = log (I_O/I)

Where I_o =P_o

and A  is the  Absorbance

    Substituting value

                A = log( (1532)/(775))

                  =0.2960

Generally beers law can be represented mathematically as

                   A = \epsilon c l

where c is the concentration of an anthracene solution

Making c the subject of the formula

          c = \frac{A}{cl}

Substituting  0.875 cm for length = b ,

We have  

            c = \frac{0.2960}{9.5*10^{3} * 0.875}

              c = 3.560 *10^{-5}M

             

6 0
3 years ago
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