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Snowcat [4.5K]
4 years ago
6

Cobalt-60 is a strong gamma emitter that has a half- life of 5.26 yr. The co balt-60 in a radiotherapy unit must be replaced whe

n its radioactivity falls to 75% of the original sample. If an original sample was purchased in June 2016, when will it be necessary to replace the cobalt-60?
Chemistry
1 answer:
Alenkasestr [34]4 years ago
3 0

<u>Answer:</u> The sample of Cobalt-60 isotope must be replaced in January 2027

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 5.26 years

Putting values in above equation, we get:

k=\frac{0.693}{5.26yrs}=0.132yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.132yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 75) = 25 grams

Putting values in above equation, we get:

0.132=\frac{2.303}{t}\log\frac{100}{25}\\\\t=10.5yrs

The original sample was purchased in June 2016

As, June is the 6th month of the year, which means the time period will be 2016+\frac{6}{12}=2016.5

Adding the time in the original time period, we get:

2016.5+10.5=2027

Hence, the sample of Cobalt-60 isotope must be replaced in January 2027

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EastWind [94]

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8. Which of the following have bacteria in their root nodules that fix nitrogen?
Natalka [10]

Answer:

Option A

Explanation:

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8 0
3 years ago
If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?
myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

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

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30.17}\ year^{-1}

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Initial concentration [A_0] = ?

Final concentration [A_t] = 12.5 grams

Applying in the above equation, we get that:-

12.5\ g=[A_0]e^{-0.02297\times 90.6}

[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g

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8 0
3 years ago
Potassium chlorate (used in fireworks, flares, and safety matches) forms oxygen and potassium chloride when heated. KClO3(s) → K
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Answer:

10.3 g of oxygen are formed when 26.4 g of potassium chlorate is heated

Explanation:

This is the balanced equation:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

Ratio beteween the salt and oxygen is 2:3

Molar mass of KClO₃ = 122.55 g/m

Let's find out the moles of salt

Mass / Molar mass

26.4 g /122.55 g/m = 0.215 moles

So, this is the final rule of three:

If 2 moles of KClO₃ make 3 moles of oxygen

0.215 moles of KClO₃ make  (0.215  .3) /2 = 0.323 moles of O₂ are produced

Molar mass O₂ = 32 g/m

Moles . molar mass = mass

0.323 m  . 32g/m = 10.3 g

4 0
3 years ago
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