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deff fn [24]
1 year ago
13

The mass of radium-226 in a sample is found to have decreased from 45.00g to 5.625g in a period of 4800 years.From this informat

ion. Calculate the half life of radium-226
Chemistry
1 answer:
yan [13]1 year ago
4 0

Answer:

Half life = 1600 years

Explanation:

Given data:

Total mass of sample = 45.00 g

Mass remain = 5.625 g

Time period = 4800 years

Half life of radium-226 = ?

Solution:

First of all we will calculate the number of half lives passes,

At time zero 45.00 g

At first half life = 45.00 g/ 2= 22.5 g

At 2nd half life = 22.5 g/ 2 = 11.25 g

At 3rd half life = 11.25 g/ 2= 5.625 g

Half life:

Half life = Time elapsed / number of half lives

Half life = 4800 years / 3

Half life = 1600 years

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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
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Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

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Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

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10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

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Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

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C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

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