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kozerog [31]
3 years ago
7

9. Thallium-208 has a half-life of 3.053 min. How long will it take for 120 g of it to decay

Chemistry
1 answer:
damaskus [11]3 years ago
4 0

Answer:

12.213 minutes will be taken for 120 g-Thalium-208 to decay to 75 grams.

Explanation:

Radioactive isotopes decay exponentially in time, the mass of the isotope (m(t)), in grams, is described by the formula in time (t), in minutes:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (1)

Where:

m_{o} - Initial mass of the isotope, in grams.

\tau - Time constant, in minutes.

In addition, the time constant associated with the isotope decay can be described in terms of half-life (t_{1/2}), in minutes:

\tau = \frac{t_{1/2}}{\ln 2} (2)

If we know that m(t) = 7.5\,g, m_{o} = 120\,g and t_{1/2} = 3.053\,min, then the time taken by the isotope is:

\tau = \frac{t_{1/2}}{\ln 2}

\tau = \frac{3.053\,min}{\ln 2}

\tau \approx 4.405\,min

t = -\tau \cdot \ln \frac{m(t)}{m_{o}}

t = -(4.405\,min)\cdot \ln \left(\frac{7.5\,g}{120\,g} \right)

t \approx 12.213\,min

12.213 minutes will be taken for 120 g-Thalium-208 to decay to 75 grams.

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<u>Answer:</u>

<em>Bronsted Lowry Theory</em>

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Bronsted Lowry Theory:

An acid is a substance that can donate one or more protons

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Hydrogen atom which is neutral (No Charge) contains 1 positive proton, 1 negative electron and 0 neutral neutron.  

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When an hydrogen atom loses an electron, Hydrogen ion is formed, which will contain 1 positive proton and 0 negative electron and 0 neutral neutron.

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Base gains H+ (proton) to form conjugate Acid

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Let us consider the example given in the question

CO_2+2H_2 O>H_3 O^+   +HCO_3^- can be written as (removed 1 H_2 O from both the sides )

CO_2+H_2 O>H^++HCO_3^- or  

CO_2+H_2 O  > H_2 CO_3 reversing the equation  

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H_2 CO_3 is the acid which donates H^+ to form HCO_3^-

H_2 O is the base which gains H^+ to form H_3 O^+

HCO_3^- is the conjugate base and H_3 O^+ is the conjugate acid

So

CO_2 is the acid

H_2 O is the base

H_3 O^+ is the conjugate acid and

HCO_3^- is the conjugate base

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