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masya89 [10]
3 years ago
5

A 200-gram sample of a radioactive nuclide is left to decay for 60 hours. At the end of 60 hours, only 25 grams of the sample is

left. What is the half-life of the nuclide sample?
60 hours
30 hours
40 hours
20 hours ---> I think 20??
Chemistry
1 answer:
Goryan [66]3 years ago
5 0
Half life is the time taken by a radioactive isotope to decay by half its original mass. 
The original mass is 200 g
Time taken is 60 hours
Final mass is 25 g
Therefore;
Final mass = Original mass × (1/2)^n; where n is the number of half lives.
25 = 200 (1/2)^n
1/8 = (1/2)^n
    n = 3 
Three half lives = 60 hours
1 half lives = 20 hours
Therefore; the half life of the radioactive nucleus is 20 hours
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<h3>Further explanation</h3>

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If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
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The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

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So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

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