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skad [1K]
3 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]3 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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vlabodo [156]

Answer:

hydrogen atom and oxygen atom

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3 years ago
If you are given a diagram of an atom, how would you identify what element it is?
Varvara68 [4.7K]

The simplest way to use the periodic table to identify an element is by looking for the element's name or elemental symbol. The periodic table can be used to identify an element by looking for the element's atomic number. The atomic number of an element is the number of protons found within the atoms of that element.





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7 0
3 years ago
Calculate the heat released when 454 grams of steam at 100 degrees C condenses.
sasho [114]
<h3>Answer:</h3>

1026.05 kJ

<h3>Explanation:</h3>
  • The Quantity of heat is calculated by multiplying the mass of a substance by specific heat capacity and the change in temperature.
  • However, condensation takes place at a constant temperature.
  • Therefore, Quantity of heat is calculated by multiplying mass by the latent heat of vaporization.
  • That is; Q = m × Lv

In this case we are given;

Mass of steam as 454 g

To calculate the amount of heat released we must know the value of latent heat of vaporization;

The latent heat of vaporization of steam is 2260 J/g

Therefore;

Heat released, Q = 454 g × 2260 J/g

                             =1026040 Joules

But, 1000 joules = 1 kilo-joules

Thus, Q = 1026.04 kJ

Hence, the amount of heat released is 1026.05 kJ

5 0
3 years ago
If 0.278g of argon dissolves in 1.5 l of water at 62 bar, what quantity of argon will dissolve at 78 bar
irinina [24]
When P1/P2 = C1/C2

and C is the molarity which = moles/volume

so, P1/P2 = [(mass1/mw)/volume] / [(mass2/mw)/volume]

P1/P2 = (mass1/mw)/1.5L / (mass2/mw)/1.5L 

so, Mw and 1.5 L will cancel out:

∴P1/P2 = mass1 / mass2

∴ mass 2 = mass1*(P2 / P1)

                = 0.278g * (78 bar / 62 bar)

                = 0.35 g

∴ the quantity of argon that will dissolve at 78 bar = 0.35 g


5 0
3 years ago
Which desribes a step in the process of forming and ionic bond
Brums [2.3K]

Explanation:

A metal atom loses electrons and becomes a positive ion.

3 0
3 years ago
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