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loris [4]
3 years ago
12

Suppose you have 100 grams of radioactive plutonium-239 with a half-life of 24,000 years. how many grams of plutonium-239 will r

emain after
a.12,000 years
b.24,000 years
c.96,000 years?
Chemistry
1 answer:
Alexandra [31]3 years ago
5 0

To solve this problem, let us first calculate for the rate constant k using the half life formula:

t1/2 = ln 2 / k

where t1/2 = half life period = 24,000 years, therefore k is:

k = ln 2 / 24,000

k = 2.89 x 10^-5 / yr

 

Now we use the rate equation:

A = Ao e^(-k t)

where,

A = mass of Plutonium-239 after number of years

Ao = initial mass of Plutonium-239

t = number of years

 

A. t = 12,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 12,000)

A = 70.7 g

 

B. t = 24,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 24,000)

A = 50 g

 

C. t = 96,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 96,000)

<span>A = 6.24 g</span>

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In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)
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<u>Answer:</u> The molarity of Cr^{3+} ions in the solution is 0.299 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For chromium (III) acetate:</u>

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol

1 mole of chromium (III) acetate (Cr(CH_3COO)_3) produces 1 mole of chromium (Cr^{3+}) ions and 3 moles of acetate (CH_3COO^-) ions

Moles of Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles

  • <u>For chromium (III) nitrate:</u>

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol

1 mole of chromium (III) nitrate (Cr(NO_3)_3) produces 1 mole of chromium (Cr^{3+}) ions and 3 moles of nitrate (NO_3^-) ions

Moles of Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles

  • <u>For chromium cation:</u>

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+} ions in the solution is 0.299 M

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