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Mariulka [41]
3 years ago
9

A radioactive isotope of potassium (k) has a half-life of 20 minutes. if a 43 gram sample of this isotope is allowed to decay fo

r 80 minutes, how many grams of the radioactive isotope will remain?
Chemistry
1 answer:
Ksenya-84 [330]3 years ago
8 0

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 43 g

m (final mass after time T) = ? (in g)

x (number of periods elapsed) = ?

P (Half-life) = 20 minutes

T (Elapsed time for sample reduction) = 80 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

80 = x*20

80 = 20\:x

20\:x = 80

x = \dfrac{80}{20}

\boxed{x = 4}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{43}{2^{4}}

m = \dfrac{43}{16}

\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

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In the compound, the sodium metal forms an ionic compound with polyatomic anion (carbonate) CO_3^{2-}.

Here, sodium is having an oxidation state of +1 called as Na^{1+} cation and carbonate CO_3^{2-} is an anion with oxidation state of -2. The charges are not balanced. So, the charges are balanced by the cris-cross method. Thus, the compound formed will be, Na_2CO_3

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Here, potassium is having an oxidation state of +1 called as K^{1+} cation and permanganate MnO_4^{1-} is an anion with oxidation state of -1. The charges are balanced. Thus, the compound formed will be, KMnO_4

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Kryger [21]

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The state equilibrium equation for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it represents a decent approximation of the activity of many gases under various conditions.

Ideal gas law can be expressed as:

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Given data:

P = 3 atm

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Putting the given data in above equation.

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