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3 years ago

The half life of radon-222 is 3.8 days. How much of a 100 gram sample is left after 15.2 days ?

Chemistry

2 answers:

Natali5045456 [20]3 years ago

5 0

Answer:

6.2 g

Explanation:

1. Calculate the number of half-lives.

2. Calculate the final mass of the substance.

nasty-shy [4]3 years ago

4 0

Answer:

6.3g of the 100g sample will be left after 15.2years

Explanation:

Half life is defined as the time taken by a radioactive substance to decay or reduce to half if its original value.

Mathematically, t1/2 = 0.693/¶ where ¶ is the decay constant.

Firstly, we need to get the decay constant using the formula for half life.

Given half life of radon (t1/2) = 3.8days

3.8 = 0.693/¶

¶ = 0.693/3.8

¶ = 0.182

If the initial value (No) if the substance is 100, the final sample (N) after 15.2years of decay can be gotten using the formula

N= Noe^-¶t where

N is the final sample after decay

No is the initial value of the sample

¶ is the decay constant

t is the time taken NY the object to decay

Given No = 100

¶ = 0.182

t = 15.2years

Substituting in the formula to get N we have;

N = 100e^-0.182(15.2)

N = 100e^-2.7664

N = 100×0.063

N = 6.3g

This shows that the initial value of radon would have decay upto 6.3g after 15.3years

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Answer:

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3 years ago

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3 years ago

Read 2 more answers

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

$24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of 0.541284 moles of MgO:

$m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}$ .

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