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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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Answer:

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Explanation:

Let us consider the complete redox reaction:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

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Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

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Zn(s) → Zn⁺²(aq) + 2e⁻

Reduction Half equation is:

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4 0
3 years ago
Calculate the final Celsius temperature when 634 L at 21 °C is compressed to 307 L.
Illusion [34]

Answer:

- 130.64°C.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have two different values of V and T:

<em>V₁T₂ = V₂T₁</em>

<em></em>

V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.

V₂ = 307.0 L, T₂ = ??? K.

<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>

<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>

5 0
3 years ago
Neptunium-237 was the first isotope of a transuranium element to be discovered. the decay constant is /s. what is the half-life
kari74 [83]

The half-life in years of Neptunium-237 which was the first isotope is 2.1 10^{6} years.

Neptunium is most stable and Neptunium-237 is undergoes alpha decay, it means Neptunium-237 is decays by the emission of alpha particles . Seven alpha particles is emitted during decay of  Neptunium-237.  Neptunium-237 is radioactive actinide  elements and first  transuranium element.

The transuranium synthesis process involves creating a transuranium element through the transmutation process . The transmutation process  is the process of creating heavy elements from light elements. Hence the process is the transmutation of light elements. There are two types: artificial and natural transmutation.

to learn more about transuranium element.

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6 0
2 years ago
Give the conjugate acid for each compound below. co3^2-
DENIUS [597]
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seraphim [82]

Answer:

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HOPE THIS HELPED

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