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3 years ago

An isotope of cesium-137 has a half-life of 30 years. If 5.0g of cesium-137 decays over 60 years. How many grams will remain.

Chemistry

1 answer:

LenKa [72]3 years ago

6 0

Answer:

1.25 gram of cesium-137 will remain.

Explanation:

Given data:

Half life of cesium-137 = 30 year

Mass of cesium-137 = 5.0 g

Mass remain after 60 years = ?

Solution:

Number of half lives passed = Time elapsed / half life

Number of half lives passed = 60 year / 30 year

Number of half lives passed = 2

At time zero = 5.0 g

At first half life = 5.0 g/2 = 2.5 g

At 2nd half life = 2.5 g/ 2 = 1.25 g

Thus. 1.25 gram of cesium-137 will remain.

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Answer:

4.17L

Explanation:

V1 = 10L

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A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill

tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

λ: is the decay constant = $ln(2)/t_{1/2}$

$t_{1/2}$ : is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

$N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg$

Now, at 7:00 pm we have:

$t = 7:00 pm - (5:00 pm + 90 min) = 30 min$

$N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg$

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

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What is the total number of oxygen atoms represented in the formula na2c03.10h2o?

Wewaii [24]

As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence

Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13

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number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23

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