<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
where,
= half life of the reaction = 5730 years
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = ? yr
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the sample = 100 grams
[A] = amount left after decay process = (100 - 25) = 75 grams
Putting values in above equation, we get:
Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %
Answer:
Explanation:
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A ratio of 0.89% w / v or 0.0089 in fraction means that
there is 0.89 weight of NaCl pero volume of solution. Therefore the mass of
NaCl is:
mass NaCl = 0.0089 * 200
<span>mass NaCl = 1.78 grams</span>
The balanced chemical equation for the combustion of butane is:
Δ
= Σ
Δ
-Σ
Δ
= ![[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]](https://tex.z-dn.net/?f=%5B%7B8%2A%28-393.5kJ%2Fmol%29%7D%2B%7B10%2A%28-241.82kJ%2Fmol%29%7D%5D-%5B%7B2%2A%28-125.6kJ%2Fmol%29%7D%2B13%2A%280%20kJ%2Fmol%29%7D%5D)
=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]
= -5315 kJ/mol
Calculating the enthalpy of combustion per mole of butane:
Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol
Correct answer: -2657.5 kJ/mol
Answer:
sediment from melting glaciers being deposited in lakes
Explanation:
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