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Andrej [43]
2 years ago
6

The half-life of radon gas is approximately four days. Four weeks after the introduction of radon into a sealed room, the fracti

on of the original amount remaining is closest to
Chemistry
1 answer:
maksim [4K]2 years ago
7 0

The fraction of the original amount remaining is closest to 1/128

<h3>Determination of the number of half-lives</h3>
  • Half-life (t½) = 4 days
  • Time (t) = 4 weeks = 4 × 7 = 28 days
  • Number of half-lives (n) =?

n = t / t½

n = 28 / 4

n = 7

<h3>How to determine the amount remaining </h3>
  • Original amount (N₀) = 100 g
  • Number of half-lives (n) = 7
  • Amount remaining (N)=?

N = N₀ / 2ⁿ

N = 100 / 2⁷

N = 0.78125 g

<h3>How to determine the fraction remaining </h3>
  • Original amount (N₀) = 100 g
  • Amount remaining (N)= 0.78125 g
  • Fraction remaining =?

Fraction remaining = N / N₀

Fraction remaining = 0.78125 / 100

Fraction remaining = 1/128

Learn more about half life:

brainly.com/question/26374513

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Answer:

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<u>Answer: </u>The mass of sample that remained is 0.127 mg

<u>Explanation:</u>

The integrated rate law equation for first-order kinetics:

k=\frac{2.303}{t}\log \frac{a}{a-x} ......(1)

Given values:

a = initial concentration of reactant = 0.500 mg

a - x = concentration of reactant left after time 't' = ?mg

t = time period = 28 s

k = rate constant = 0.049s^{-1}

Putting values in equation 1:

0.049s^{-1}=\frac{2.303}{28s}\log (\frac{0.500}{(a-x)})\\\\\log (\frac{0.500}{(a-x)})=\frac{0.049\times 28}{2.303}\\\\\frac{0.500}{a-x}=10^{0.5957}\\\\frac{0.500}{a-x}=3.94\\\\a-x=\frac{0.500}{3.942}=0.127mg

Hence, the mass of sample that remained is 0.127 mg

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A sample of an unknown liquid has a volume of 30.0 mL and a mass of 6 g. What is its density? i need this hella quick
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density= mass/volume

density= 6g/30.0 ml

density= 0,2 g/ml

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