Hopefully this will help you my friend.
Answer:
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
<em>Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.</em>
We can find rate constant from half-life as follows:
<em>Rate constant:</em>
t(1/2) = ln 2 / K
As half-life of Cesium-137 is 30.2 years:
30.2 years = ln 2 / K
<em>K = 0.02295 years⁻¹</em>
Replacing this result and with the given data of the problem:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.02295 years⁻¹* t + ln[A]₀
Ln ([A] / [A₀]) = -0.02295 years⁻¹* t
<em>As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:</em>
Ln (0.2) = -0.02295 years⁻¹* t
70.1 years = t
<h3>There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value</h3>
methanol:
1 mole CH3 OH --> produces --> 1 mole CO2
1 mole CO2 has a molar mass of 44.01 gh/mole
your set up is:
(44.01 g CO2) / -726.5kJ = 0.06058g
your answer 0.06058 grams of CO2 produced per kJ released.
Answer:
A. Add a coefficient 4 before h2o and add a 2 before so2
Explanation:
