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 = N / N₀
Fraction remaining = 0.78125 / 100
Fraction remaining = 1/128
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I am pretty sure that the answer is B. extrusive rock. I hope that this helps you!! Good Luck!!
Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr
